在Java中获得超时错误

Question

我正在进行谷歌挑战，我必须解决一个难题。对于这个难题，我必须测试在任何四边触摸它的任何数字是否等于一。如果是这样，我必须改变相反的方向。如果一个则为零。如果为零。当我尝试运行以下代码时，我试图运行它的代码实验室说下面的＆＃34; Code Timed Out＆＃34;。我查了一下，d看起来循环应该是无限的。无论它最终应该返回-1还是步数。任何人都可以解释这一点。

import java.util.Arrays;

public class FooReal {

    public static boolean[] elementsOn = {false, false};

    public static void main(String args[]) {
        int[][] grid = {{1, 1}, {0, 0,}};
        System.out.println(answer(grid));
    }

    public static int answer(int[][] grid) {
        int steps = 0;

        while (containsOn(grid)) {
            for (int index = 0; index < grid.length; index++) {
                for (int subIndex = 0; subIndex < grid[index].length; subIndex++) {
                    if (grid[index][subIndex] == 1) {
                        elementsOn = isElementOn(grid, new int[] {index, subIndex});
                        if (elementsOn[0] || elementsOn[1]) {
                            steps++;
                            for (int counter = 0; counter < grid.length; counter++) {
                                if (grid[counter][subIndex] == 1) {
                                    grid[counter][subIndex] = 0; 
                                } else {
                                    grid[counter][subIndex] = 1;
                                }
                            }

                            for (int counterTwo = 0; counterTwo < grid[index].length; counterTwo++) {
                                if (grid[index][counterTwo] == 1) {
                                    grid[index][counterTwo] = 0;
                                } else {
                                    grid[index][counterTwo] = 1;
                                }
                            }

                            if (grid[index][subIndex] == 1) {
                                grid[index][subIndex] = 0;
                            } else {
                                grid[index][subIndex] = 1;
                            }
                        }
                    }
                }

                if (steps > 15) {
                    steps = -1;
                    return steps;
                }

            }
        }
        return steps;
    }

    public static boolean[] isElementOn(int[][] grid, int[] elements) {
        boolean[] elementsOn = {false, false};

        //Check if any are on in x row.
        try {
            if (grid[elements[0] + 1][elements[1]] == 1) {
                elementsOn[0] = true;
            }
        } catch (ArrayIndexOutOfBoundsException e) {}
        try {
            if (grid[elements[0] - 1][elements[1]] == 1) {
                elementsOn[0] = true;
            }
        } catch (ArrayIndexOutOfBoundsException e) {}

        //Check if any are on in y row.
        try {
            if (grid[elements[0]][elements[1] + 1] == 1) {
                elementsOn[1] = true;
            }
        } catch (ArrayIndexOutOfBoundsException e) {}
        try {
            if (grid[elements[0]][elements[1] - 1] == 1) {
                elementsOn[1] = true;
            }
        } catch (ArrayIndexOutOfBoundsException e) {}

        return elementsOn;
    }

    public static boolean containsOn(int[][] grid) {
        for (int index = 0; index < grid.length; index++) {
            for (int subIndex = 0; subIndex < grid[index].length; subIndex++) {
                if (grid[index][subIndex] == 1) {
                    return true;
                }
            }
        } return false;
    }

}